Question

A four digit number (numbered from 0000 to 9999) is said to be lucky if the sum of its first two digits is equal to the sum of its last two digits. If a four digit number is picked at random, then find the probability that it is a lucky number.
(a) 0.07
(b) 0.067
(c) 0.67
(d) 0.08

Answer 1

First, before proceeding for this, we must know the method of writing the four digits number whose digits at thousand place is a, hundred place is b, tens place is c and ones place is d as 1000(a)+100(b)+10(c)+d. Then, the condition given in the question which says it is lucky when the sum of its first two digits is equal to the sum of its last two digits is $a+b=c+d$. Then, we know the formula to calculate the sum of series of square of individual terms is given by $\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{n}^{2}} \right)=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$and by solving it, we get the final answer.

Complete step by step answer:
In this question, we are supposed to find the probability that a number be lucky when a four digit number (numbered from 0000 to 9999) is said to be lucky if the sum of its first two digits is equal to the sum of its last two digits and is selected randomly.
So, before proceeding for this, we must know the method of writing the four digits number whose digits at thousand place is a, hundred place is b, tens place is c and ones place is d as:
1000(a)+100(b)+10(c)+d
Now, the condition given in the question which says it is lucky when sum of its first two digits is equal to the sum of it last two digits as:
$a+b=c+d$
Now, let us suppose (a+b) as n and then we make different combinations for the sum of the numbers as:
For n=0, we get the total number of conditions as 1 only as only a number 0000 comes in this category.
For n=1, we have combinations as 1010, 0110, 1001 and 0101 which is 4 means as a square of 2.
For n=2, we have combinations as 0202, 0220, 0211, 2002, 2020, 2011, 1102, 1120 and 1111 which are 9 in total means square of 3.
Similarly, we can conclude that for n=3, total cases are 16 which is square of 4 and so on.
Now, the last number 9999 gives the total of 1 case with n=18.
Also, we have noticed that the series of squares goes till 9 which is 81 and it is covering two times.
So, we get the total numbers which are lucky is given by the sum of series as:
$2\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+{{6}^{2}}+{{7}^{2}}+{{8}^{2}}+{{9}^{2}} \right)+{{10}^{2}}$
Then, we know the formula to calculate the sum of series of square of individual terms is given by:
$\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{n}^{2}} \right)=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
So, after substituting the value of n as 9 as it reaches the square of 8 at maximum, we get:
$\begin{aligned} & 2\times \dfrac{\left( 9 \right)\left( 9+1 \right)\left( 2\times 9+1 \right)}{6} \\\ & \Rightarrow \dfrac{\left( 9 \right)\left( 10 \right)\left( 19 \right)}{3} \\\ & \Rightarrow 3\times 10\times 19 \\\ & \Rightarrow 570 \\\ \end{aligned}$
Also, we have one condition in which we get the sum as 9 we get a square of 10 which is 100.
So, we get the total number of lucky cases as:
570+100=670
Now, we also can see clearly that total numbers are from 0000 to 9999 which are 10,000 in total.
So, to get the probability of selecting lucky number, we get:
$\begin{aligned} & \dfrac{670}{10000} \\\ & \Rightarrow 0.067 \\\ \end{aligned}$
So, we get the probability as 0.067.

So, the correct answer is “Option B”.

Note: Now, to solve these types of the questions we need to know some of the basics of the probability to get the final answer. So, probability is defined as the ratio of the favourable outcomes to the total outcomes. Also, in this question, we must be very careful as we get only one case for getting the square of 10 when the sum is 9.